最新修改合肥八中高二第一学期期末考试卷

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2022-2023学年安徽省合肥八中高二(上)期末化学模拟试卷

2022-2023学年安徽省合肥八中高二(上)期末化学模拟试卷

2022-2023学年安徽省合肥八中高二(上)期末化学模拟试卷一.选择题(共16小题,满分18分)1.下列有关能量的判断或表示方法正确的是()A.从C(s,石墨)═C(s,金刚石)△H=+1.9 kJ•mol﹣1,可知石墨比金刚石更稳定B.等质量的硫蒸气和硫固体分别完全燃烧,后者放出热量更多C.由H+(aq)+OH﹣(aq)═H2O(l)△H=﹣57.3 kJ•mol﹣1,可知含1 mol CH3COOH的溶液与含1 mol NaOH的溶液混合,放出热量等于57.3 kJD.2gH2完全燃烧生成液态水放出285.8 kJ热量,则氢气燃烧的热化学方程式为2H2(g)+O2(g)═2H2O (l)△H=﹣285.8kJ•mol﹣12.(3分)如图为铜铁原电池示意图,心里有关说法正确的是()A.电子由铁棒通过导线流向铜棒B.正极反应为Fe﹣2e﹣═Fe2+C.铜棒逐渐溶解D.该装置能将电能转化为化学能3.已知:2A(g)⇌B(g)+Q (Q>0);2A(g)⇌B(l);下列能量变化示意图正确的是()A.B.C.D.4.(3分)I2在KI溶液中存在下列平衡:I2(aq)+I﹣(aq)⇌I3﹣(aq)某I2、KI混合溶液中,I3﹣的物质的量浓度c(I3﹣)与温度T的关系如图所示(曲线上任何一点都表示平衡状态,忽略I2的挥发).下列说法正确的是()A.反应I2(aq)+I﹣(aq)⇌I3﹣(aq)△H>0B.状态A与状态B相比,状态A的c(I2)大C.若反应进行到状态D时,一定有v逆>v正D.若温度为T1、T2,反应的平衡常数K1、K2则K1>K25.采用阴离子交换法合成了一系列不同Zn 和Pt 含量的PtSn﹣Mg(Zn)AlO 催化剂用于乙烷脱氢反应[CH3CH3(g )⇌CH2=CH2(g)+H2(g)△H>0],实验结果表明,在水滑石载体中掺杂少量的Zn 对乙烷脱氢反应有明显影响,如图所示为不同Zn 含量PtSn 催化剂的乙烷催化脱氢反应中,乙烷的转化率随时间的变化。

安徽省合肥市一中、六中、八中2023-2024学年高二化学第一学期期末监测模拟试题含解析

安徽省合肥市一中、六中、八中2023-2024学年高二化学第一学期期末监测模拟试题含解析

安徽省合肥市一中、六中、八中2023-2024学年高二化学第一学期期末监测模拟试题注意事项1.考生要认真填写考场号和座位序号。

2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。

第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。

3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。

一、选择题(共包括22个小题。

每小题均只有一个符合题意的选项)1、下列做法符合“绿色化学”思想的是A.回收利用废旧电池B.在通风橱中制取SO2不用NaOH溶液吸收尾气C.将实验室的有机废液直接倒入下水道D.用燃气热水器代替太阳能热水器2、分别将下列物质:①CuCl2②Mg(HCO3)2③Na2SO3④NH4HCO3⑤KMnO4溶于水中,对其溶液加热蒸干后并灼烧,仍能得到原物质的是()A.①③B.②⑤C.③D.无3、一定条件下反应2AB(g)A2(g)+B2(g)达到平衡状态的标志是A.容器内压强不随时间变化B.容器内,3种气体AB、A2、B2共存C.容器中各组分的体积分数不随时间变化D.AB的消耗速率等于A2的消耗速率4、对于可逆反应N2(g)+3H2(g) 2NH3(g) ΔH<0,下列各项对示意图的解释与图像相符的是A.①压强对反应的影响(p2>p1)B.②温度对反应的影响C.③平衡体系增加N2对反应的影响D.④催化剂对反应的影响5、对于平衡体系:aA(g)+bB(g)cC(g)+dD(g) △H<0,下列判断其中不正确的是A.若温度不变,容器体积扩大一倍,此时A的浓度是原来的0.45倍,则a+b<c+dB.若从正反应开始,平衡时,A、B的转化率相等,则投入A、B的物质的量之比为a:bC.若平衡体系中共有气体Mmol,再向其中充入bmolB,达到平衡时气体总物质的量为( M+b)mol,则a+b=c+d D.若a+b=c+d,对于体积不变的容器,升高温度,平衡向左移动,容器中气体的压强不变6、如图表示1~18号元素原子的结构或性质随核电荷数递增的变化。

合肥市第六中学、第八中学、168中学等校2021-2022学年高二上学期期末考试英语试题(含答案)

合肥市第六中学、第八中学、168中学等校2021-2022学年高二上学期期末考试英语试题(含答案)

合肥八中等学校2021~2022学年度第一学期高二期末联考英语本试卷由四个部分组成。

其中,第一、二部分和第三部分的第一节为选择题。

第三部分的第二节和第四部分为非选择题。

考试时间120分钟,满分150分。

考生注意事项:1. 答题前,考生务必在试题卷、答题卡规定的地方填写自己的姓名、座位号。

2. 回答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

回答非选择题时,将答案写在答题卡上,写在本试卷上无效。

3. 考试结束,务必将试题卷和答题卡一并上交。

第一部分听力(共两节,满分30分)回答听力部分时,请先将答案标在试卷上。

听力部分结束前,你将有两分钟的时间将你的答案转涂到客观题答题卡上。

第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. 【此处可播放相关音频,请去附件查看】What colour shirt did Gracie decide on?A. Red.B. White.C. Black.2. 【此处可播放相关音频,请去附件查看】What does the man think of Professor Green’s course?A. Boring.B. Useless.C. Easy.3. 【此处可播放相关音频,请去附件查看】What time is it now?A. 4:50.B. 4:30.C. 4:20.4. 【此处可播放相关音频,请去附件查看】Where does the conversation probably take place?A. At a gas station.B. At an art gallery.C. In a workshop.5. 【此处可播放相关音频,请去附件查看】What is the man supposed to do?A. Meet the headmaster.B. Borrow some books.C. Attend a meeting.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

安徽省合肥市第八中学2021-2022学年高二上学期期末考试化学试题含答案

安徽省合肥市第八中学2021-2022学年高二上学期期末考试化学试题含答案

2021~2022学年度第一学期高二期末考试化学考生注意:1.本试卷分选择题和非选择题两部分.满分100分,考试时间90分钟.2.答题前,考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚. 3.考生作答时,请将答案答在答题卡上.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题区域书写的答案无效.............,在试题卷....、草稿纸上....作答无效..... 4.本卷命题范围:人教版选择性必修1.5.可能用到的相对原子质量:H1C12O16Na23S32一、选择题(本大题共16小题,每小题3分,共计48分.在每小题列出的四个选项中,只有一项是符合题目要求的)1.下列说法正确的是()A .化学键断裂放出能量B .所有化学反应均伴随能量变化C .液体凝固吸收热量D .总焓:若反应物>生成物,则0H ∆>2.某同学以柠檬为材料自制水果电池,装置如图,下列叙述正确的是()A .H +由镁极区移向银极区B .镁片上有无色气体放出 C .铜片替代镁片,现象相同D .银片为负极 3.已知:1()() P s,P s,21.7kJ mol H -∆=-⋅红磷黑磷;1P s,P s,17.6kJ mol ()() H -∆-⋅=白磷红磷.由此推知,其中最稳定的磷单质是()A .红磷B .白磷C .黑磷D .无法确定4.25℃时,在某恒容密闭烧瓶中存在如下平衡:()()2242NO g N O g 0H ∆<.若把烧瓶置于100℃的水中,则烧瓶内气体属性不会改变的是() A .颜色B .平均分子量C .压强D .密度5.将()amol X g 和()bmol Y g 通入密闭容器中发生反应()()()aX g bY g cZ g +,通过实验得到不同条件下达到平衡时Z 的物质的量分数与温度和压强的关系如图.下列关于该反应的H S ∆∆、的判断正确的是()A .0,0H S ∆>∆>B .0,0H S ∆<∆<C .0,0H S ∆>∆<D .0,0H S ∆<∆> 6.苹果酸为二元有机弱酸(2H A ).下列关于2H A 及2Na A 叙述正确的是() A .电离方程式:22H A2H A +-+B .2H A 溶液中含有2种分子:2H A 与2H OC .2Na A 溶液:()()+c H c OH ->D .2Na A 溶液:()()c OH c HA --<7.山东舰是我国自主研制的新型航母,为了延长航母服役寿命可以在航母舰体(主要成分是钢铁合金)上镶嵌金属锌.下列有关说法正确的是()A .可以用铅等金属代替锌B .构成原电池反应时,舰体表面发生氧化反应C .这种保护方法叫牺牲阳极的阴极保护法D .在酸雨环境中,航母主要发生吸氧腐蚀8.一定条件下,向某密闭容器中充入等物质的量的X 和Y ,发生反应:()()()X g 2Y g 2Z g +.达到平衡后,测得反应前气体的总物质的量与反应后气体的总物质的量之比为5:4,则X 的转化率为() A .80%B .66%C .40%D .33%9.甲胺(32CH NH )为一元有机弱碱,其电离及与酸反应类似于3NH ,下列关于稀甲胺水溶液的叙述错误的是()A .其他条件不变,加水稀释,()33c CH NH +变大B .其他条件不变,升温,促进了32CH NH 的电离C .电离方程式:32233CH NH H OCH NH OH +-++D .与稀盐酸反应化学方程式:3233CH NH HCl CH NH Cl +10.在绝热恒容反应器中发生反应()()()()223SO g NO g SO g NO g ++,反应速率随时间的变化如图.下列叙述正确的是()A .3t 时反应达到平衡状态B .该反应为吸热反应C .34t t :v 正~因温度降低而降低D .4t 时2SO 的转化率达到最大 11.焊接时常用4NH Cl 除锈,下列说法一定正确的是() A .4NH Cl 为弱酸强碱盐B .4NH Cl 可以抑制水的电离C .pH 7=的4NH Cl 、氨水混合液:()()4c NH c Cl+-> D .4NH Cl 溶液:()()()32c H c OH c NH H O +-=+⋅ 12.已知下列热化学方程式:()()()()221C s H O l CO g H g H ++∆()()()2222CO g O g 2CO g H +∆;()()()22232H g O g 2H O 1 H +∆()C s 的燃烧热为()A .123H H H ∆+∆∆—B .1231122H H H ∆+∆+∆ C .1232H H H ∆+∆+∆D .12312H H H ∆+∆+∆13.一定温度下,在1L 密闭容器中,将CO 和2H S 混合加热并达到下列平衡:()()()()22CO g H S g COS g H g 0.1 K ++=.反应前CO 物质的量为5mol ,平衡后CO物质的量为4mol .则反应前2H S 的物质的量浓度为()A .13.5mol L -⋅B .13mol L -⋅C .12.5mol L -⋅D .11.25mol L -⋅ 14.粗银的精炼工艺原理如图,下列叙述错误的是()A .y 极电极材料为粗银B .x 极主要电极反应式为Ag e Ag -+-C .装置乙将化学能转化为电能D .n 极电极反应式为22O 4H 4e 2H O +-++15.硒酸钡(4BaSeO )为特种玻璃的添加剂,制取原理为2244Ba SeO BaSeO +-+↓,在4BaSeO 饱和溶液中()2+lgc Ba 与()24lgc SeO --关系如图所示(4BaSeO 溶于水时吸收热量).下列叙述错误的是()A .升温:M 点上移B .X 点:214a c(SeO )10mol L ---=⋅C .M 点:21c(Ba )1mol L +-=⋅D .sp K :M N X ==点点点 16.室温下,下列关于电解质溶液的说法中不正确的是()A .将pH 12=的氨水与pH 2=的硝酸等体积混合:()()()()43c NH c NO c OH c H +--+>>> B .室温下,调节3H A 溶液的pH 5=时,20.63c(HA )10c(H A)-=已知:2.27.212.4a1a2a3K 10,K 10K 10---===、)C .25℃时,22RS M MS R ++++的平衡常数K 4000=[该温度下,()21sp K MS 1.010-=⨯,()18sp K RS 4.010-=⨯]D .加水稀释氯化铝溶液,铝离子水解程度增大,H +物质的量增加,溶液酸性增强二、非选择题(本题共5小题,共52分)17.(10分)氢元素单质及其化合物是人类赖以生活的重要能源.回答下列问题 (1)H H,OO H O —、—的键能分别为111436kJ mol 496kJ mol 463kJ mol ---⋅⋅⋅、、. ①热化学方程式()()()2222H g O g 2H O g , H H +∆∆为___________1kJ mol -⋅.②氢气的燃烧热1285.8kJ mol H -∆⋅=-,表示氢气燃烧热的热化学方程式为_______________________③1mol 水蒸气液化放热____________kJ .(2)肼(24N H )是一种液态火箭推进剂.24N H 分解的能量变化如图所示:①正反应的活化能为__________1kJ mol -⋅,气态肼分解的热化学方程式为______________________②该反应的S ∆_______(填“>”“<”或“=”)0,该反应自发进行的条件为____________(填“高温”“低温”或“任意温度”). 18.(10分)电化学是当今化学研究的热点之一,回答下列问题:(1)碱性电池具有耐用、电流量大,储存寿命长、外壳不易腐蚀等优点.铝一铜碱性原电池的工作原理如装置甲所示:①在__________(“Al ”或“Cu ”)极有2H 生成,Al 极电极反应式为_________________________ ②若有33.6L (标准状况下)2H 生成,则理论上电解质溶液增重___________g .(2)某兴趣小组设计以甲烷、氧气酸性燃料电池为电源电解NaCl 溶液,并验证氯气的某种化学性质,工作原理如装置乙、装置丙所示. ①m 极为_________极,装置乙发生反应的化学方程式为____________________________________ ②Y 极电极反应式为__________________________________③气球b 中的现象为_______________________________________________________________,证明了氯气的化学性质之一:_______________________性.19.(12分)一定条件下,CO 与水蒸气反应生成2CO 和2H 的热化学方程式为()()()()1222 CO g H O g CO g H g 41.3kJ mo l H -++∆⋅=-.回答下列问题:(1)已知1mol C 完全转化成水煤气(2CO H 、混合物)时吸收热量131.5kJ ,则1mol C 与2CO 反应生成CO 的焓变是____________________(2)上述反应达到平衡的标志是___________________(填字母). a .容器内压强不随时间变化b .容器内各物质的浓度不随时间变化c .容器内2CO H O 、的浓度之比为1∶1d .单位时间消耗0.1mol CO 同时生成20.1mol H O(3)研究表明:上述反应平衡常数随温度的变化如下表所示:温渡/℃ 400 500 800 平衡常数K1595若反应为500℃,起始时CO 和2H O 的浓度均为1mo 002l .L -⋅,则达到平衡时,CO 的平衡转化率为____________(4)在T 5MPa ℃、条件下,将0.1mol CO 与()20.12mol H O g 混合气体充入1L 密闭容器发生反应,反应过程中2CO 的物质的量浓度随时间的变化如图所示:①在02min ~内,以CO 表示的平均反应速率为___________11mol L min --⋅⋅.②T ℃时该反应的平衡常数p K =____________(压强代替浓度,分压=总压×物质的分数). ③平衡后,若向容器中再充入0.2mol CO 和()20.02mol H O g ,则再次平衡后容器内的压强与充入CO 和()2H O g 前的压强比为_____________20.(9分)硫代硫酸钠(223Na S O )又称大苏打,可用作定影剂、还原剂.现有某种硫代硫酸钠样品,为了测定该样品纯度,某兴趣小组设计如下实验方案,回答下列问题:(1)溶液配制:称取10.0g 该硫代硫酸钠样品,用_________________________(填“新煮沸并冷却的蒸馏水”或“自来水”)在烧杯中溶解,完全溶解后,冷却至室温,再用玻璃棒引流将溶液全部转移至100mL 的容量瓶中,加蒸馏水至离容量瓶刻度线12cm ~处,改用_______________定容,定容时,视线与凹液面相切.(2)滴定:取10.10mol L -⋅的227K Cr O (硫酸酸化)标准溶液20.00mL ,加入过量KI ,发生反应:232722Cr O 6I 14H 3I 2Cr 7H O --++++++.然后加入淀粉溶液作为指示剂,用硫代硫酸钠样品溶液滴定至终点,发生反应:2222346I 2S O S O 2I ---++,重复实验,平均消耗223Na S O 样品溶液的体积为20.0mL .①取用硫酸酸化的227K Cr O 标准溶液可选用的仪器是__________(填“甲”或“乙”). ②滴定终点时溶液的颜色变化是______________________(3)样品纯度的计算:①根据上述有关数据,该样品中223Na S O 的质量分数为____________________②下列操作可能使测量结果偏低的是________(填字母). A .盛装硫代硫酸钠样品溶液的滴定管没有润洗 B .锥形瓶中残留少量水C .读数时,滴定前仰视,滴定后俯视 21.(11分)氨水及铵盐是化学工业的重要原料.回答下列问题:(1)氨水中各种离子的物质的量的浓度从小到大的顺序是____________,其他条件不变,往稀氨水溶液中加入少量蒸馏水,则混合液中水的电离程度____________(填“增加”“减小”或“不变”).(2)常温下向某氨水中缓缓通入HCl 气体,溶液中432c(NH )lg c(NH )H O +⋅与pH 的关系如图所示.常温下32NH O H ⋅的电离平衡常数b K =_______(用含有a 的代数式表示),温度升高A 点的位置可能是__________(填“A1”或“A2”),a_______ (填“>”“<”或“=”)7.(3)某化学兴趣小组以某废液(主要含有3+3+2+Al Fe Mn 、、)为原料制取3MnCO 的工艺流程及各离子开始沉淀、完全沉淀的pH 如图所示(常温下3MnCO 的11sp K 910-=⨯):金属离子 3+Fe3+Al 2+Mn开始沉淀pH2.2 4.1 8.8完全沉淀pH3.55.410.8①“沉铁、铝”时反应液的pH 应控制的范围是______________________ ②“沉锰”时温度不能较高的原因是________________________,2+Mn完全沉淀时()251c Mn 10mol L +--<⋅,则此时溶液中()23c CO ->___________1mol L -⋅.2021~2022学年度第一学期高二期末考试·化学参考答案,提示及评分细则1.B2.A3.C4.D5.B6.B7.C8.C9.A10.D11.D 12.B13.A14.A15.C16.D 17.(1)①484-(1分)②12221H (g)O (g)H O(1) . 2858kJ mol 2H -∆⋅+=-(2分)③43.8(2分)(2)①a 50.7+(1分);()()()12422 N H g N g 2H g 50.7kJ l mo H -+∆⋅=+(2分)②>:高温(各1分)18.(1)①Cu (1分);22Al 4OH 3e AlO 2H O ---+-+(2分)②24(2分)(2)①正;4222CH 2O CO 2H O ++(各1分)②22Cl 2e Cl ---↑(1分)③气球b 中充满黄绿色气体,淀粉碘化钾试纸变蓝:氧化(各1分) 19.(1)1172.8kJ mol -⋅+(2)bd (3)75%(4)①0.01②0.05③2∶1(每空2分) 20.(1)新煮沸并冷却的蒸馏水;胶头滴管(各1分) (2)①甲(1分)②溶液蓝色消失且半分钟内不恢复原色(2分) (3)①94.8%(2分)②A (2分)21.(1)()()()4c OH c NH c H-++>>(2分);增大(1分) (2)1410a -(2分);A2(1分);>(1分) (3)①[)5.4,8.8或其他合理表示形式(1分) ②防止铵盐分解及氨气挥发(1分);6910-⨯(2分)。

2020-2021学年安徽省合肥一中、六中、八中联考高二(上)期末物理试卷

2020-2021学年安徽省合肥一中、六中、八中联考高二(上)期末物理试卷

2020-2021学年安徽省合肥一中、六中、八中联考高二(上)期末物理试卷一、单选题(本大题共6小题,共24.0分)1.如图所示,固定的水平长直导线中通有电流I,矩形线框与导线在同一竖直平面内,且一边与导线平行.线框由静止释放,在下落过程中()A. 穿过线框的磁通量保持不变B. 线框中感应电流方向保持不变C. 线框所受安培力的合力为零D. 线框的机械能不断增大2.如图所示,三根长直导线平行固定在空间中,其截面构成等边三角形。

若仅在a、b中通入大小均为I的电流,方向如图所示,此时a受到的磁场力大小为F.然后又在c中通入如图所示方向的电流,a受到的磁场力大小仍为F,此时下列说法中正确的是()A. b受到的磁场力大小为√3FB. b受到的磁场力的方向平行于ac的连线C. ac连线中点d的磁场方向垂直于ac指向bD. c中的电流强度为2I3.两电荷量分别为q1和q2的点电荷固定在x轴上的A、B两点,两电荷连线上各点电势ϕ随坐标x变化的关系图象如图所示,其中P点电势最高,且x AP<x PB用,则()A. q1和q2都是正电荷B. q1的电荷量大于q2的电荷量C. 在A、B之间将一负点电荷沿x轴从P点左侧移到右侧,电势能先减小后增大D. 一点电荷只在电场力作用下沿x轴从P点运动到B点,加速度逐渐变小4.在图示电路中,A、B为两块正对的水平金属板,G为静电计。

开关S闭合后,静电计指针张开一个角度,板间的带电油滴悬浮在两板之间静止不动。

下列说法正确的是()A. 若仅将A板竖直向下缓慢平移一些,则静电计指针的张角将减小B. 若仅将A板竖直向下缓慢平移一些,则油滴将向下运动C. 若断开S,且仅将A板竖直向下缓慢平移一些,则油滴将向上运动D. 若断开S,且仅将A板水平向右缓慢平移一些,则油滴将向上运动5.在如图所示的电路中,当开关S置于a处时,电流表(内阻不计)示数为I,额定功率为16W的电动机正常工作,带动质量为0.7kg的物体以2m/s的速度匀速竖直上升。

安徽省合肥市2023-2024学年高二上学期期末考试数学试题含答案

安徽省合肥市2023-2024学年高二上学期期末考试数学试题含答案

2023-2024学年第一学期高二年级期末检测数学试题卷(答案在最后)注意事项:1.你拿到的试卷满分为150分,考试时间为150分钟.2.试卷包括“试题卷”和“答题卷”两部分,请务必在“答题卷”上答题,在“试题卷”上答题无效.第Ⅰ卷(选择题)一、选择题:本题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合题目要求的.1.在数列{}n a 中,11111n n a a a +==+,,则4a =()A.2B.32 C.53D.85【答案】C 【解析】【分析】由数列的递推公式,依次求出234,,a a a 即可.【详解】数列{}n a 中,11111n na a a+==+,,则有21112a a =+=,321312a a =+=,431513a a =+=.故选:C.2.“26m <<”是“方程22126x y m m+=--表示的曲线为椭圆”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【答案】B 【解析】【分析】利用椭圆的标准方程结合充分、必要条件的定义计算即可.【详解】易知26m <<时,20,60m m ->->,但4m =时有262m m -=-=,此时方程表示圆,所以不满足充分性,若方程22126x ym m +=--表示的曲线为椭圆,则()()20602,44,626m m m m m->⎧⎪->⇒∈⋃⎨⎪-≠-⎩,显然26m <<成立,满足必要性,故“26m <<”是“方程22126x y m m+=--表示的曲线为椭圆”的必要不充分条件.故选:B3.已知直线60x ay -+=和直线()3230a x y a ++-=互相平行,则实数a 的值为()A.1-或2B.1-或2- C.2- D.1-【答案】D 【解析】【分析】根据平行关系列式求a 的值,并代入检验即可.【详解】由题意可得:()32a a -+=,解得1a =-或2a =-,若1a =-,则两直线分别为60,2230x y x y ++=++=,符合题意;若2a =-,则两直线均为260x y ++=,不符合题意;综上所述:1a =-.故选:D.4.已知等差数列{}n a 的前n 项和为n S ,且36430a S ==,,则4a =()A.2- B.2C.4D.6【答案】D 【解析】【分析】根据等差数列的性质和前n 项求和公式计算即可求解.【详解】由题意知,616346()3()302S a a a a =+=+=,又34a =,所以43106a a =-=.故选:D5.已知x a =是函数21()(1)ln 2f x x a x a x =-++的极大值点,则实数a 的取值范围是()A.(,1)-∞B.(1,)+∞ C.(01),D.(0,1]【答案】C 【解析】【分析】求导后,得导函数的零点,1a ,比较两数的大小,分别判断在x a =两边的导数符号,确定函数单调性,从而确定是否在x a =处取到极大值,即可求得a 的范围.【详解】21()(1)ln 2f x x a x a x =-++,则()()1()(1)x a x a f x x a x x--=-++=',0x >,当(0,1)a ∈时,令()0f x '>得0x a <<或1x >,令()0f x '<得1<<a x ,此时()f x 在区间(0,)a 上单调递增,(),1a 上单调递减,()1,+∞上单调递增,符合x a =是函数()f x 的极大值点;当1a =时,()21()0x f x x-'=≥恒成立,函数()f x 不存在极值点,不符合题意;当(1,)a ∞∈+时,令()0f x '>得01x <<或x a >,令()0f x '<得1x a <<,此时()f x 在区间(0,1)上单调递增,()1,a 上单调递减,(),a +∞上单调递增,符合x a =是函数()f x 的极小值点,不符合题意;综上,要使函数()f x 在x a =处取到极大值,则实数a 的取值范围是(01),.故选:C.6.从某个角度观察篮球(如图1)可以得到一个对称的平面图形(如图2),篮球的外轮廓为圆O ,将篮球的表面粘合线视为坐标轴和双曲线,若坐标轴和双曲线与圆O 的交点将圆的周长八等分,且||||||AB BC CD ==,则该双曲线的离心率为()A.43B.167C.7D.97【答案】C 【解析】【分析】设双曲线的标准方程为()222210,0x y a b a b-=>>,求出圆O 与双曲线在第一象限内的交点E 的坐标,将点E 的坐标代入双曲线的方程,可得出ba的值,再利用双曲线的离心率公式可求得该双曲线的离心率.【详解】设双曲线的标准方程为()222210,0x y a b a b-=>>,设圆O 与双曲线在第一象限内的交点为E ,连接DE 、OE ,则33==+==OE OD OC CD OC a,因为坐标轴和双曲线与圆O 的交点将圆O 的周长八等分,则1π2π84DOE ∠=⨯=,故点,22⎛⎫⎪ ⎪⎝⎭E ,将点E的坐标代入双曲线的方程可得2222221⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭-=a b ,所以2297b a =,所以,该双曲线的离心率为7ce a===.故选:C.7.如图,在三棱锥A BCD -中,1,AD CD AB BC AC =====,平面ACD ⊥平面ABC ,则三棱锥A BCD -外接球的表面积为()A.3πB.8π3C.7π3D.2π【答案】B 【解析】【分析】先确定底面ABC 的外接圆圆心,结合图形的特征,利用勾股定理及外接球的表面积公式计算即可.【详解】如图所示,取AC 中点E ,连接,DE BE ,在BE 上取F 点满足2EF FB =,由题意易知ABC 为正三角形,则F 点为ABC 的外接圆圆心,且,ED AC BE AC ⊥⊥,因为平面ACD ⊥平面ABC ,平面ACD 平面ABC AC =,所以DE ⊥底面ABC ,BE ⊥底面ADC ,过F 作//FO DE ,故三棱锥A BCD -外接球的球心O 在直线FO 上,作OG EF //交DE 于G 点,设OF h =,球半径为R ,根据1,AD CD AB BC AC =====易知,,2263BE AE DE EF BF =====,四边形OGEF 为矩形,由勾股定理可知:222222OB OF BF OD OG DG =+==+,即22222120,3263R h h h R ⎛⎛⎫=+=-+⇒== ⎪ ⎪⎝⎭⎝⎭,故其外接球表面积为28π4π3S R ==.故选:B8.已知0.98ln 0.98a =-,1b =, 1.02 1.02ln1.02c =-,则()A.a b c <<B.c b a <<C.b<c<aD.b a c<<【答案】B 【解析】【分析】利用()ln ,0f x x x x =->的单调性可判断a b >,利用()ln (0)g x x x x x =->的单调性可判断c b <,故可得三者之间的大小关系.【详解】设()ln ,0f x x x x =->,则有11()1x f x x x'-=-=,∴当01x <≤时,()()0,f x f x '≤在(]0,1上单调递减;(0.98)(1)1f f ∴>=,即有0.98ln 0.981->,a b ∴>;令()ln (1)g x x x x x =-≥,则()1(ln 1)ln g x x x '=-+=-,∴当1x ≥时,()0g x '≤,当且仅当1x =时等号成立,故()g x 在[)1,∞+上单调递减;(1.02)(1)1g g ∴<=,即有1.02 1.02ln1.021-<,c b ∴<,综上所述,则有c b a <<,故选:B.二、选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求,全部选对的得5分,部分选对的得2分,有选错的得0分.9.已知直线():20R l ax y a a ++-=∈与圆22:5C x y +=,则下列结论正确的是()A.直线l 必过定点B.l 与C 可能相离C.l 与C 可能相切D.当1a =时,l 被C 截得的弦长为【答案】ACD 【解析】【分析】利用直线方程确定过定点可判定A ,利用直线与圆的位置关系可判定BC ,利用弦长公式可确定D.【详解】由直线方程变形得()():120l a x y -++=,显然1x =时=2y -,即直线过定点()1,2-,故A 正确;易知()22125+-=,即点()1,2-在圆C 上,则直线l 不会与圆相离,但有可能相切,故B 错误,C 正确;当1a =时,此时直线:10l x y ++=,圆心为原点,半径为r =,则圆心到l 的距离为d =,所以l 被C 截得的弦长为=,故D 正确.故选:ACD10.定义:设()f x '是()f x 的导函数,()f x ''是函数()f x '的导数,若方程()0f x ''=有实数解0x ,则称点()()0x f x ,为函数()y f x =的“拐点”.经探究发现:任何一个三次函数都有“拐点”且“拐点”就是三次函数图像的对称中心.已知函数()321533f x x ax bx =+++的对称中心为()1,1,则下列说法中正确的有()A.1,0a b =-= B.函数()f x 既有极大值又有极小值C.函数()f x 有三个零点 D.对任意x ∈R ,都有()()11f x f x -+=【答案】AB 【解析】【分析】根据拐点定义二次求导可计算可求出函数解析式即可判定A ,根据导数研究其极值可判定B ,结合B 项结论及零点存在性定理可判定C ,利用函数解析式取特殊值可判定D.【详解】由题意可知()22f x x ax b '=++,()22f x x a ''=+,而()()151113301022f a b a b f a⎧==+++=-⎧⎪⇒⎨⎨=⎩⎪==+⎩'',故A 正确;此时()321533f x x x =-+,()()222f x x x x x '=-=-,显然2x >或0x <时,()0f x ¢>,则()f x 在()(),0,2,-∞+∞上单调递增,()0,2x ∈时,()0f x '<,即()f x 在()0,2上单调递减,所以()f x 在0x =时取得极大值,在2x =时取得极小值,故B 正确;易知()()()5100,250,2033f f f =>-=-<=>,结合B 结论及零点存在性定理可知()f x 在()2,0-存在一个零点,故C 错误;易知()()510113f f +=+≠,故D 错误.故选:AB11.如图,已知抛物线()220C y px p =>:的焦点为F ,抛物线C 的准线与x 轴交于点D ,过点F 的直线l (直线l 的倾斜角为锐角)与抛物线C 相交于A B ,两点(A 在x 轴的上方,B 在x 轴的下方),过点A 作抛物线C 的准线的垂线,垂足为M ,直线l 与抛物线C 的准线相交于点N ,则()A.当直线l 的斜率为1时,4AB p =B.若NF FM =,则直线l 的斜率为2C.存在直线l 使得AOB 90∠=D.若3AF FB =,则直线l 的倾斜角为60【答案】AD 【解析】【分析】根据抛物线的焦点弦的性质一一计算即可.【详解】易知,02p F ⎛⎫⎪⎝⎭,可设():02p AB y k x k ⎛⎫=-> ⎪⎝⎭,设()()1122,,,A x y B x y ,与抛物线方程联立得()22222220242p y k x k p k x k p p x y px⎧⎛⎫=-⎪ ⎪⇒-++=⎝⎭⎨⎪=⎩,则221212224k p p p x x x x k ++==,,对于A 项,当直线l 的斜率为1时,此时123x x p +=,由抛物线定义可知12422p pAF BF x x AB p +=+++==,故A 正确;易知AMN 是直角三角形,若NF FM =,则ANM FMN AMF FAM ∠=∠⇒∠=∠,又AF AM =,所以AMF 为等边三角形,即60AFx ∠= ,此时3k =B 错误;由上可知()()222212121212124pk p k x x y y k x x x x +=+-++()()2222222223104244p k p pk p k k p k +=+⨯-⨯+=-<,即0OA OB ×<uu r uu u r,故C 错误;若1212332322p p AF FB x x x p x ⎛⎫=⇒-=-⇒=- ⎪⎝⎭ ,又知212213,462p p px x x x =⇒==,所以1y =,则112y k p x ==-,即直线l 的倾斜角为60 ,故D 正确.故选:AD12.如图,在棱长为2的正方体1111ABCD A B C D -中,已知,,M N P 分别是棱111,,C D AA BC 的中点,Q 为平面PMN 上的动点,且直线1QB 与直线1DB 的夹角为30 ,则()A.1DB ⊥平面PMNB.平面PMN 截正方体所得的截面图形为正六边形C.点Q 的轨迹长度为πD.能放入由平面PMN分割该正方体所成的两个空间几何体内部(厚度忽略不计)的球的半径的最大值为32【答案】ABD 【解析】【分析】A 选项,建立空间直角坐标系,求出平面PMN 的法向量,得到线面垂直;B 选项,作出辅助线,找到平面截正方体所得的截面;C 选项,作出辅助线,得到点Q 的轨迹,并求出轨迹长度;D 选项,由对称性得到平面PMN 分割该正方体所成的两个空间几何体对称,由对称性可知,球心在1B D 上,设球心坐标建立方程,求出半径的最大值.【详解】A 选项,如图所示以D 为坐标原点,建立空间直角坐标系,则()()()()11,2,0,0,1,2,2,0,1,2,2,2P M N B ,故()()()12,2,2,1,1,2,1,2,1DB PM PN ==--=-.设平面PMN 的法向量为(),,m x y z = ,则2020m PM x y z m PN x y z ⎧⋅=--+=⎪⎨⋅=-+=⎪⎩,令11z x y =⇒==得()1,1,1m =,易知12DB m =,故1DB ⊥平面PMN ,即A正确;B 选项,取111,,AB CC AD 的中点,,F QE ,连接11,,,,,,,,NE NF ME MQ PQ PF A B EP D C ,结合题意可知11////,////NF A B EP EP CD MQ ,所以N F P E 、、、四点共面且M Q P E 、、、四点共面,两个平面都过点P ,所以M Q P E N F 、、、、、六点共面,易知EM MQ QP PF FN NE ======,所以平面PMN 截正方体所得的截面为正六边形ENFPQM ,B正确;C 选项,由上知1DB ⊥平面PMN ,设垂足为S ,以S 为圆心133B S 为半径在平面PMN 上作圆,由题意可知Q 轨迹即为该圆,结合B 的结论可知平面PMN 平分正方体,根据正方体的中心对称性可知S 平分1DB,故半径1111332B S DB =⨯=,故点Q 的轨迹长度为2π,C 错误;D 选项,由上知该两部分空间几何体相同,不妨求能放入含有顶点D 的这一空间几何体的球的半径最大值,结合A 项空间坐标系及正方体的对称性知该球球心O 在1DB 上,该球与平面PMN 切于点S ,与平面ABCD 、平面11A D DA 、平面11D C CD 都相切,设球心为()(),,01O a a a a <≤,则球半径为a ,易知()1,1,1S ,故()223312RS a a a a -=⇒-=⇒=,D 正确.故选:ABD 【点睛】思路点睛:关于立体几何中截面的处理思路有以下方法(1)直接连接法:有两点在几何体的同一个平面上,连接该两点即为几何体与截面的交线,找截面就是找交线的过程;(2)作平行线法:过直线与直线外一点作截面,若直线所在的平面与点所在的平面平行,可以通过过点找直线的平行线找到几何体与截面的交线;(3)作延长线找交点法:若直线相交但在立体几何中未体现,可通过作延长线的方法先找到交点,然后借助交点找到截面形成的交线;(4)辅助平面法:若三个点两两都不在一个侧面或者底面中,则在作截面时需要作一个辅助平面.关于立体几何中求动点轨迹的问题注意利用几何特征,比如动直线与定直线夹角为定值,可以考虑结合圆锥体得出动点轨迹.第Ⅱ卷(非选择题)三、填空题:本题共4小题,每小题5分,共20分.13.已知正方体1111ABCD A B C D -的棱长为a ,则异面直线1A B 与1B C 所成的角的余弦值_________________.【答案】12##0.5【解析】【分析】利用正方体的特征构造平行线求异面直线夹角即可.【详解】如图所示连接1,A D BD ,根据正方体的特征易知11//B C A D ,且1A DB △为等边三角形,所以1DA B ∠即异面直线1A B 与1B C 所成的角,且160DA B ∠= ,11cos 2DA B ∠=.故答案为:1214.在正项等比数列{}n a 中,若234234111502a a a a a a ++=++=,,3a =_____________.【答案】5【解析】【分析】根据正项等比数列的定义与通项公式,计算即可【详解】正项等比数列{}n a 中,23450a a a ++=,234242334332224323234343323111502a a a a a a a a a a a a a a a a a a a a a a a ++++++++====,解得35a =±,舍去负值,所以35a =.故答案为:515.以两条直线1220350l x y l x y +=++=:,:的交点为圆心,并且与直线3490x y -+=相切的圆的方程是_____________________.【答案】()()221216x y -++=【解析】【分析】直接利用交点坐标和点到直线的距离公式求出圆心和半径,最后求出圆的方程.【详解】利用20350x y x y +=⎧⎨++=⎩,解得12x y =⎧⎨=-⎩,则圆心坐标为()1,2-,设圆的方程为()()22212x y r -++=利用圆心()1,2-到直线3490x y -+=的距离d r =,整理得4r ==,故圆的方程为()()221216x y -++=.故答案为:()()221216x y -++=.16.关于x 的不等式()1e ln x a x x a x +--≥恒成立,则实数a 的最大值为_____________________.【答案】2e 2【解析】【分析】构造函数()()e 1ln ,xf x x xg x x=+-=,利用导数研究其单调性及最值,分离参数计算即可.【详解】设()()()e 1ln 0,xf x x x xg x x=+->=,易知()()()2e 11,x x x f x g x x x''--==,则当1x >时,()()0,0f x g x ''>>,即此时两函数均单调递增,当01x <<时,()()0,0f x g x ''<<,即此时两函数均单调递减,故()()()()12,1e f x f g x g ≥=≥=,对于不等式()()11ln e ln e 1ln x x x a x x a a x x x++---≥⇔≥+-,由上可知1ln 2u x x =+-≥,故1ln e 1ln x xa x x+-≤+-,又()()e 2u g u u u =≥单调递增,故()()2e 22g u g a ≥=≥.所以实数a 的最大值为2e 2.故答案为:2e 2.【点睛】关键点点睛:观察不等式结构可发现是指对同构式即原式等价于()1ln e 1ln x x a x x +-≥+-,构造函数()()e 1ln ,xf x x xg x x=+-=判定其单调性与最值分参计算即可.四、解答题:本题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.已知数列{}n a 满足()111,211n n a a a n n n n +-==++.(1)证明数列{}n na 为等差数列,并求出数列{}n a 的通项公式;(2)设21n nb n a =,数列{}n b 的前n 项和为n S ,求20S .【答案】(1)证明见解析,1+=n n a n (2)202021S =【解析】【分析】(1)根据题中递推公式可得()111n n n a na ++-=,结合等差数列的定义和通项公式分析求解;(2)由(1)可得111n b n n =-+,利用裂项相消法运算求解.【小问1详解】因为()1111n n a a n n n n +-=++,则()111n n n a na ++-=,所以数列{}n na 是以首项112a ⨯=,公差1d =的等差数列,可得211n n na n =+-=+,所以1+=n n a n .【小问2详解】由(1)可得()2111111n n b n a n n n n ===-++,所以20111111201122320212121S =-+-+⋅⋅⋅+-=-=.18.设圆C 与两圆()()22221221,21C x y C x y ++=-+=::中的一个内切,另一个外切.(1)求圆心C 的轨迹E 的方程;(2)已知直线()00x y m m -+=>与轨迹E 交于不同的两点,A B ,且线段AB 的中点在圆2210x y +=上,求实数m 的值.【答案】(1)2213y x -=(2)2±【解析】【分析】(1)根据圆与圆的位置关系结合双曲线的定义分析求解;(2)联立方程结合韦达定理运算求解.【小问1详解】圆()22121C x y ++=:的圆心为()12,0C -,半径为1,圆()22221C x y -+=:的圆心为()22,0C ,半径为1,设圆C 的半径为r ,若圆C 与圆1C 内切,与圆2C 外切,则121,1CC r CC r =-=+,可得212CC CC -=;若圆C 与圆2C 内切,与圆1C 外切,则211,1CC r CC r =-=+,可得122CC CC -=;综上所述:122CC CC -=,可知:圆心C 的轨迹E 是以1C 、2C 为焦点的双曲线,且1,2a c ==,可得2223b c a =-=,所以圆心C 的轨迹E 的方程2213y x -=.【小问2详解】联立方程22130y x x y m ⎧-=⎪⎨⎪-+=⎩,消去y 得222230x mx m ---=,则()()222Δ4831220m m m =---=+>,可知直线与双曲线相交,设()()1122,,,A x y B x y ,线段AB 的中点为()00,M x y,可得120003,222x x m m x y x m +===+=,即3,22m m M ⎛⎫ ⎪⎝⎭,且3,22m m M ⎛⎫ ⎪⎝⎭在圆2210x y +=上,则2231022m m ⎛⎫⎛⎫+= ⎪ ⎪⎝⎭⎝⎭,解得2m =±,所以实数m 的值为2±.19.如图所示,用平面11BCC B 表示圆柱的轴截面,BC 是圆柱底面的直径,O 为底面圆心,E 为母线1CC 的中点,已知1AA 为一条母线,且14AB AC AA ===.(1)求证:平面AEO ⊥平面1AB O ;(2)求平面1AEB 与平面OAE 夹角的余弦值.【答案】(1)证明见解析;(2)6.【解析】【分析】(1)根据图形特征结合勾股逆定理先证11,B O AO B O EO ⊥⊥,由线线垂直得线面垂直,根据线面垂直的性质可得面面垂直;(2)建立合适的空间直角坐标系,求出平面的法向量和平面的法向量,利用向量法能求出二面角的余弦值.【小问1详解】依题意可知AB AC ⊥,则ABC 是等腰直角三角形,故AO BC ⊥,由圆柱的特征可知1BB ⊥平面ABC ,又AO ⊂平面ABC ,1BB AO ⊥,因为11,BB BC B BB BC =⊂ 、平面11BCC B ,则AO ⊥平面11BCC B ,而1B O ⊂平面11BCC B ,则AO ⊥1B O ,因为14AB AC AA ===,则2221124BC B O B B BO ==∴=+=,222222*********,36OE OC CE B E E C B C B O OE =+==+==+,所以1B O OE ⊥,因为1B O OE ⊥,AO ⊥1B O ,,AO OE O AO OE =⊂ 、平面AEO ,所以1B O ⊥平面AEO ,因为1B O ⊂平面1AB O ,所以平面AEO ⊥平面1AB O ;【小问2详解】由题意及(1)知易知1,,AA AB AC 两两垂直,如图所示建立空间直角坐标系则()()()14,0,4,0,4,2,2,2,0B E O ,所以()()()114,0,4,0,4,2,2,2,4AB AE B O ===-- ,由(1)知1B O 是平面AEO 的一个法向量,设(),,n x y z = 是平面1AB E 的一个法向量,则有1440420n AB x z n AE y z ⎧⋅=+=⎪⎨⋅=+=⎪⎩ ,取22,1z x y =-⇒==,所以()2,1,2n =- ,设平面1AEB 与平面OAE 的夹角为θ,所以111cos cos ,6n B O n B O n B Oθ⋅====⋅ .即平面1AEB 与平面OAE夹角的余弦值为6.20.已知函数()ln ,f x a x x a =-∈R .(1)设1x =是()f x 的极值点,求a 的值,并求()f x 的单调区间;(2)证明:当2a ≤时,()10f x x+<在()1,+∞上恒成立.【答案】(1)1a =,单调区间见解析(2)证明见解析【解析】【分析】(1)求导,根据极值的定义分析求解,进而可得单调区间;(2)根据题意分析可得()112ln f x x x x x +<-+,令()12ln ,1g x x x x x =-+>,利用单调性判断其单调性和符号,即可得结果.【小问1详解】因为()ln f x a x x =-的定义域为()0,∞+,则()1a f x x'=-,若1x =是()f x 的极值点,则()110f a -'==,解得1a =,当1a =,则()ln f x x x =-,()111x f x x x-=-=',令()0f x '>,解得01x <<;令()0f x '<,解得1x >;则()f x 在()0,1内单调递增,在()1,∞+内单调递减,可知1x =是()f x 的极大值点,即1a =符合题意,所以()f x 的单调递增区间为()0,1,单调递减区间为()1,∞+.【小问2详解】因为()1,x ∞∈+,则ln 0x >,且2a ≤,可得ln 2ln a x x ≤,即()112ln f x x x x x+≤-+,令()12ln ,1g x x x x x =-+>,则()()22212110x g x x x x-=--=-<'在()1,∞+内恒成立,可知()g x 在()1,∞+内单调递减,可得()()10g x g <=,即()112ln 0f x x x x x +≤-+<,所以当2a ≤时,()10f x x +<在()1,∞+上恒成立.21.对每个正整数(),,n n n n A x y 是抛物线24x y =上的点,过焦点F 的直线n FA 交抛物线于另一点(),n n n B s t .(1)证明:()41n n x s n =-≥;(2)取12n n x +=,并记n n n a A B =,求数列{}n a 的前n 项和.【答案】(1)证明见解析(2)11142134n n n +⎛⎫-+- ⎪⎝⎭【解析】【分析】(1)设直线:1n n n y A k B x =+,联立方程结合韦达定理分析证明;(2)根据抛物线的定义结合(1)可得1424n n n a =++,利用分组求和法结合等比数列求和公式运算求解.【小问1详解】由题意可知:抛物线24x y =的焦点()0,1F ,且直线n n A B 的斜率存在,设直线:1n n n y A k B x =+,联立方程214n y k x x y=+⎧⎨=⎩,消去y 得2440n x k x --=,可得216160n k ∆=+>,所以()41n n x s n =-≥.【小问2详解】因为12n n x +=,由(1)可得142242n n n n s x +=-=-=-,则22144144,44444n n n n nn n n x s y t +======,可得12424n n n n n n n a A B y t ==++=++,设数列{}n a 的前n 项和为n T ,则()21221114442444n n n n T a a a n ⎛⎫=++⋅⋅⋅+=++⋅⋅⋅++++⋅⋅⋅++ ⎪⎝⎭()1111414441124211143414n nn n n n +⎡⎤⎛⎫-⎢⎥ ⎪-⎝⎭⎢⎥⎛⎫⎣⎦=++=-+- ⎪-⎝⎭-,所以11142134n n n T n +⎛⎫=-+- ⎪⎝⎭.【点睛】关键点点睛:利用韦达定理证明关系,并根据抛物线的定义求n a .22.已知椭圆()222210+=>>x y C a b a b :的离心率32,点3⎛ ⎝⎭在椭圆上.(1)求椭圆C 的方程;(2)设点()()()()0,1,,0,4,02A M t N t t -≠,直线AM AN ,分别与椭圆C 交于点,(,S T S T 异于),A AH ST ⊥,垂足为H ,求OH 的最小值.【答案】(1)2214x y +=(221-【解析】【分析】(1)根据题意结合离心率列式求,,a b c ,进而可得方程;(2)联立方程求,S T 的坐标,根据向量平行可知直线ST 过定点()2,1Q ,进而分析可知点H 在以AQ 为直径的圆上,结合圆的性质分析求解.【小问1详解】由题意可得:2222213142a b c a b c e a ⎧⎪=+⎪⎪+=⎨⎪⎪==⎪⎩,解得21a b c ⎧=⎪=⎨⎪=⎩,所以椭圆C 的方程为2214x y +=.【小问2详解】由题意可得:直线:AM x ty t =-+,联立方程2214x ty t x y =-+⎧⎪⎨+=⎪⎩,消去x 可得()22224240t y t y t +-+-=,解得2244t y t -=+或1y =,可知点S 的纵坐标为2244t t -+,可得2224844t t x t t t t -=-⋅+=++,即22284,44t t S t t ⎛⎫- ⎪++⎝⎭,同理可得:()()()()2228444,4444t t T t t ⎛⎫--- ⎪ ⎪-+-+⎝⎭,即()22284812,820820t t t T t t t t ⎛⎫--+ ⎪-+-+⎝⎭,取()2,1Q ,则()222228,44t QS t t ⎛⎫- ⎪=-- ⎪++⎝⎭ ,()222228,820820t QT t t t t ⎛⎫- ⎪=-- ⎪-+-+⎝⎭,因为()()222222222288082044820t t t t t t t t ⎡⎤⎡⎤--⎛⎫⎛⎫-----=⎢⎥⎢⎥ ⎪ ⎪-+++-+⎝⎭⎝⎭⎢⎥⎢⎥⎣⎦⎣⎦,可知QS ∥QT ,即,,Q S T 三点共线,可知直线ST 过定点()2,1Q ,又因为AH ST ⊥,且()0,1A ,可知:点H 在以AQ 为直径的圆上,该圆的圆心为()1,1E ,半径112r AQ ==,所以OH的最小值为1OE r -=.。

安徽省合肥市第八中学2021-2022学年高二上学期期末考试生物试题含答案

安徽省合肥市第八中学2021-2022学年高二上学期期末考试生物试题含答案
C.激素通过体液运输而神经递质不通过体液传递
D.体液调节相对于神经调节来说作用范围更广,作用时间更长
【答案】C
8.如图表示人体血糖平衡调节的部分过程示意图,下列叙述正确的是()
A.血糖平衡的调节中枢位于下丘脑
B.激素Z能够促进肝糖原和肌糖原的分解
C.激素Z 胰高血糖素,通过导管分泌到体液中
D.激素Z和胰岛素为协同作用,共同维持人体血糖平衡
B.利用高浓度2,4-D作除草剂,可抑制农田中杂草的生长
C.用生长素类似物处理未授粉的番茄花蕾,可得到无籽番茄
D.植物激素较人工合成的植物生长调节剂作用效果更稳定
【答案】D
15.植物的生长发育是多种激素相互作用、共同调节的结果。为了探究赤霉素和生长素对植物生长的影响,科研人员将大小、生理状况等相同的豌豆幼茎切段若干,随机分为两部分,进行了两组实验,实验设计及实验结果如下表所示。下列推断错误的是()
A.图一中的a、b、c、d、e共同构成内环境
B.图二中的甲相当于图一中的b,不属于内环境的组成部分
C.图二中的乙相当于图一中的a,内含营养物质和代谢废物
D.图一中组织细胞代谢过旺时可能导致过程3强于4引发组织水肿
【答案】C
2.某同学通过比较自来水,缓冲液和生物材料在加入酸或碱后pH的变化,推测生物体是如何维持pH稳定的。部分实验步骤如下:
D.神经递质作用于突触后膜,引起下一个神经元兴奋
【答案】B
4.反射是在中枢神经系统参与下,机体对内外环境刺激所作出的规律性反应。婴儿时期就可完成一些简单的非条件反射,之后由于机体不断接触外界环境,经过训练可逐渐形成越来越多的条件反射。下列相关叙述正确的是()
A.在反射过程中,兴奋在神经元间单向传递,在神经纤维上双向传导

安徽省合肥八中2024年高二物理第一学期期末学业质量监测试题含解析

安徽省合肥八中2024年高二物理第一学期期末学业质量监测试题含解析
故选B
4、B
【解题分析】A.在0<t<2×10-10s时间内,Q板比P板电势高5V,所以电场方向水平向左,电子所受电场力方向向右,加速度方向也向右,所以电子从M点向右做匀加速直线运动;故A错误.
B.在2×10-10s<t<4×10-10s时间内,Q板比P板电势低5V,电场强度方向水平向右,所以电子所受电场力方向向左,加速度方向也向左,所以电子向右做匀减速直线运动,当t=4×10-10s时速度为零,此时电子在M点 右侧;故B正确.
【题目详解】A、带电粒子的轨迹向左弯曲,则带电粒子所受的电场力沿电场线切线向左,与电场线的方向相反,知带电粒子带负电;故A错误.
B、电场线的疏密表示场强大小,由图知粒子在B点的场强小于A点的场强,在B点的加速度小于A点的加速度;故B错误.
C、从A到B,电场力做负功,动能减小,速度也减小,故带电粒子在A点的速度大于在B点的速度;故C正确.
安徽省合肥八中2024年高二物理第一学期期末学业质量监测试题
注意事项:
1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。用2B铅笔将试卷类型(B)填涂在答题卡相应位置上。将条形码粘贴在答题卡右上角"条形码粘贴处"。
2.作答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。答案不能答在试题卷上。
D.磁场中某点的磁感应强度是由磁场本身决定的,与该点是否放通电导线无关,选项D正确。
故选D。
6、D
【解题分析】要扩大电流变量程,应该并联分流电阻,利用并联电路的规律和欧姆定律可以求出分流电阻的大小
【解题分析】牛顿运动定律能够解决宏观物体的低速运动问题,在生产、生活及科技方面起着重要作用;解决问题时虽然有一定误差,但误差极其微小,可以忽略不计;故经典力学仍可在一定范围内适用.虽然相对论和量子力学更加深入科学地认识自然规律,它是科学的进步,但并不表示对经典力学的否定,故选项B正确.A、C错误;经典力学不能用于处理高速运行的物体;故D错误

2024届安徽合肥八中物理高二第一学期期末达标检测模拟试题含解析

2024届安徽合肥八中物理高二第一学期期末达标检测模拟试题含解析
B.电压表(0~15V内阻约20k Ω)
C.电流表(0~0.3A,内阻约1 Ω)
D.电流表(0~0.6A,内阻约0.4 Ω)
E.滑动变阻器(5 Ω,1A)
F.滑动变阻器(500 Ω,0.2A)
G.电源(电动势3V,内阻1 Ω)
(1)实验中电压表应选用_____________,电流表应选用_____________,为使实验误差尽量减小,要求电压表从零开始变化且多取几组数据,滑动变阻器应选用___________(用序号字母表示)
【题目点拨】本题也可以求解出以20m/s的速度转弯时所需的向心力,与将侧向最大静摩擦力与所需向心力比较,若静摩擦力不足提供向心力,则车会做离心运动
6、D
【解题分析】根据波形图和Q点的振动图像读出波长和周期,求解波的速度;根据Q点t=0.1s的振动方向判断波的传播方向;根据波的传播规律判断在t=0.4 s时刻质点P的振动情况以及t=0.7 s时刻质点Q的振动情况.
(1)小球到达C点的速度大小
(2)小球在C点时,轨道受到的压力大小
15.(12分)如图,两根间距为L=0.5m的平行光滑金属导轨间接有电动势E=3V、内阻r=1Ω的电源,导轨平面与水平面间的夹角θ=37°.金属杆ab垂直导轨放置,质量m=0.2kg.导轨与金属杆接触良好且金属杆与导轨电阻均不计,整个装置处于竖直向上的匀强磁场中.当R0=1Ω时,金属杆ab刚好处于静止状态,取g=10m/s2,sin37°=0.6,cos37°=0.8
故选D
【题目点拨】关键是知道汽车在水中的倒影相对于汽车静止不动
5、D
【解题分析】汽车转弯时做圆周运动,重力与路面的支持力平衡,侧向静摩擦力提供向心力,根据牛顿第二定律分析解题
【题目详解】汽车转弯时受到重力,地面的支持力,以及地面给的摩擦力,其中摩擦力充当向心力,A错误;当最大静摩擦力充当向心力时,速度为临界速度,大于这个速度则发生侧滑,根据牛顿第二定律可得 ,解得 ,所以汽车转弯的速度为20m/s时,所需的向心力小于1.4×104N,汽车不会发生侧滑,BC错误;汽车能安全转弯的向心加速度 ,即汽车能安全转弯的向心加速度不超过7.0m/s2,D正确

2021-2022学年安徽省合肥市八中高二上学期1月期末考试英语试题及答案

2021-2022学年安徽省合肥市八中高二上学期1月期末考试英语试题及答案

2021-2022学年安徽省合肥市八中高二上学期1月期末考试英语试题★祝考试顺利★(含答案)本试卷分选择题和非选择题两部分。

满分150分"考试时间120分钟。

本卷命题范围:北师大版选择性必修第一册Unit 3〜选择性必修第三册Unit 7。

第一部分听力(共两节,满分30分)第一节(共5小题;每小题1. 5分,满分7. 5分)听下面5段对话。

每段对话后有一个小题.从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What are the speakers talking about?A. Tennis.B. Tom.C. Homework.2. What did the woman order yesterday?A. Some paper.B. A printer.C. Forms.3. What is the man doing?A. Taking photos.B. Drawing a picture.C. Encouraging the woman.4. What does the woman mean?A. The man will meet many people.B. The man plans to buy some food.C. The man will have to clean the house.5. What will the woman do?A. Make a phone call.B. Check the ATM.C. Go to the bank.第二节(共15小题;每小题1.5分,满分22. 5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

安徽巢湖一中、合肥八中2021-2022学度高二上学期年末联考--英语

安徽巢湖一中、合肥八中2021-2022学度高二上学期年末联考--英语

安徽巢湖一中、合肥八中2021-2022学度高二上学期年末联考--英语第一部分听力(共两节,满分30)第一节(共5小题,每小题1.5分,满分7.5)()1.When will Linda come?A.Today. B.Tomorrow. C.Next week.()2.Where is Paul now?A.At school B.At home.C.In the hospital.()3.How will the woman go to the airport?A.By taxi. B.By bus.C.By underground.()4.How much does the woman weigh now?A.150 pounds. B.155 pounds.C.160 pounds.()5.What is the woman’s hobby?A. Watching TV. B.Going to the movies. C.Taking pictures.二、(共15小题,每小题1.5分,满分22.5分)听第6段对话,回答第6和第7两个小题。

()6.Who do you think Jack probably is?A.A tourist. B.A Londoner. C.A performer.()7 . Why won’t they see Swan Play together?A.They haven’t booked the seats ye t.B.Jack will have to leave.C.They don’t like it.听第7段对话,回答第8至第10三个小题。

()8.What is the probable relationship between the two speakers?A.Strangers.B.Teacher and student.C.Husband and wife.()9.Why do people in England like to talk about the weather? A.Because it is interesting.B.Because it’s a safe topic.C.Because the weather is changeable.()10.What may happen if strangers talk about politics according to the man?A.They may have a disagreement.B.They may be interested in it.C.They may stop their talk soon.听第8段对话,回答第11至第13三个小题。

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最新修改合肥八中高二第一学期期末考试卷9.Why is the man going to Beijing?A. To travel..B. To find a jobC.To do business.10.How will the man go to Beijing?A.By train.B.By bus.C.By car.11.What does the man ask the woman to do?A.Get his things ready.B.Buy him a ticket.C.Wish him good luck.听下面一段对话,回答第12至第14三个小题。

12.What are the two speakers talking about?A.The Browns' trip to the USA.B.The Browns' plan for the trip to China.C.The Browns' trip to China.13.What did the Browns suggest the two speakers should do?A.Take warm clothes there.B.Take less clothes there.C.Not buy souvenirs there.14.What is cheap in China according to the conversation?A.Clothes.B.Souvenirs .C.Gold jewelry.15.What does the woman want to do?A.Promote products.B.Keep fit.C.Buy gym equipment.16.By whom will a plan be made?A.The womanB.The trainer. .C.The man.17.How much will the service charge?A.300 yuan a week.B.400 yuan a month.C.800 yuan a year.听下面一段独白,回答第18至第20三个小题。

18.Who would prefer to eat the roasted meat?A.People in India.B.People in the desert.C.Eskimos in the North Pole.19.In which parts of the world is coffee very popular?A.In Northern Europe and in the Middle East.B.In North America and in the east.C.In Southern Europe and in the Middle East.20.How do people in England like drinking tea?A.With nothing.B.With sugar,milk or lemons.C.With sugar,cold milk or coffee.二.单项选择(共15小题,每小题1分,满分15分)21. Everything comes with _____ price; there is no such ______ thing as free lunch in the world.A. a;aB. the;/C. the;aD. a;/22. I tried to work on,but I was so tired that I could no longer ________.A.bear out B.hold on C.hold up D.work out 23.I strongly recommend that the information ________ in my report ________ to Mr. Brown without delay.A.to be referred to;to be e-mailed B.referred to;e-mailC.referred to;be e-mailed D.being referred to;being e-mailed24. The difficulty the driver _______ driving on the road in the morning was that he _________ a thick fog.A. had; ran outB. had; ran intoC. met; ran out ofD. met; ran into25. Much new and high technology has been introduced form America, thus _________ a great increase in the production of the company.A. resulting inB. resulted fromC. results inD. results from26. ________ a reply from the company, he decided to write another letter to defend his legal rights.A. Not to have receivedB. Not having receivedC. Not to receiveD. Having not received27. He is widely ________ to be one of the best player in the world.A. acknowledgedB. expandedC. clarifiedD. applauded28. As John Lennon once said, life is _______ happens to you while you are busy making other plans.A. whichB. thatC. whatD. where29. We still haven’t ________ how to predict the earthquake.A. got throughB. come upC. got overD. figured out30. After ten days of travelling in Africa, ________, the blacks came back to their homeland.A. tired and happilyB. tired but happyC. tiredly and happilyD. tiredly but happy31. One minute she burst into _______ and the next minute burst out ________. We just couldn’t catch her mood at any moment.A. tears; laughingB. crying; laughterC. tears; laughterD. crying; laughing32. The appearance of Howard and Harden has _______ the fans of a sure victory for the Huston Rocket in the coming NBA contest.A. accusedB. expectedC. convincedD.reminded33. Nowadays, almost all the Chinese people are _________ to basic health care service. This accounts for the fact that the average life of the Chinese has already risen to 75.A.a lternativeB. abundantC.accessible D. creative34. It was the boy _________ had been in prison _________ stole the money.A. who, whereB. that, howC. that, whichD. who, that35. --- Could I use this dictionary?--- _________. It’s a spare one.A. Good ideaB. Just go aheadC. You’re welcomeD. You’d better not二.完形填空(共20小题,每小题1分,满分20分)A Good Friend, A Second SelfI couldn’t believe my ears when I heard my name called for the leading role in our high school play. Mrs. Dermitt, my drama teacher,had been looking for someone to play an energetic boy in a comedy. Luckily for me, she thought that I could handle the 36 .That afternoon my friend Kevin and I talked 37 about the play. Although Kevin hadn’t been 38 for a part onstage, his job with the set crew was important to the success of the play.I told him I was a little 39 because I had a lot of lines to memorize.“You can do it.” He said. I knew I could 40 him: we had been friends since the third grade, and we 41 a good team.Preparations for the play moved at a rapid pace. While working hard with the set crew, Kevin 42 spent hours helping me learn my lines. He often said my lines with me by silently moving his lips. We 43 that he could probably play my part as well as I could.Three days before the 44 night, everything was ready for the performance. But when I woke up 45 a fever and sore throat on the day of the play, the entire productioncame to a sudden 46 . Everyone in the drama department was worried, 47 there was no way I could perform. The play was 48 to open in fewer than six hours, and we had no time to cancel. I tried to think of a way to 49 . Then it hit me –Kevin knew the 50 as well as I did. I called Mrs. Dermitt to give her my 51 . Within a few short hours, Kevin stood onstage in costume and makeup. The amusing lines he had 52 with me so many times made the crowd laugh and cheer. In a strange turn of events, Kevin and I had 53 the day for everyone by working as a team.Of course, I was terribly disappointed to have 54 my chance in the spotlight, but I was extremely 55 to have such a good friend.36.A.part B.play C.band D.crew 37.A.calmly B.seriouslyC.excitedly D.anxiously 38.A.elected B.chosen C.invitedD.trained39.A.upset B.bored C.nervousD.confused40.A.depend on B.tend to C.talk withD.agree with41.A.joined B.found C.tookD.made42.A.also B.only C.still D.simply 43.A.expected B.debated C.agreedD.joked44.A.final B.special C.opening D.greeting 45.A.with B.from C.in D.by 46.A.change B.turn C.stopD.end47.A.for B.but C.and D.so 48.A.likely B.ready C.sureD.due49.A.pass B.help C.care D.face 50.A.steps B.lines C.pointD.case51.A.introduction B.instructionC.explanationD.suggestion52.A.scanned B.graspedC.practiced D.presented 53.A.valued B.saved C.leftD.kept54.A.offered B.avoided C.riskedD.missed55.A.successful B.thankfulC.trustful D.hopeful 三.阅读理解(共20小题,每小题2分,满分40分)AEnglish is one of the most widely spoken languages in the world. Learning English well is very important. Here are some methods that can help you learn English well.Do a lot of listening. Listen to recordings of your teacher,the radio,TV video cassettes audio cassettes一anything that you can get your hands on. Just listen to English as much as you can and you will learn a lot.Every time you learn something new,write it down in a notebook. This will help youremem-ber what you have learned. Buy a notebook and use it only for practicing English.Try to read English one hour every day. Yes,this is difficult,but English has many words,and a good way to enlarge your vocabulary is to read them.Whenever you have the chance,speak English. This will be hard to do if you don’t live inan English-speaking country; however, the Internet provides new chances to practice speakingEnglish through chat rooms and audio-video chat technology. You can also record your voice using a recorder.Find a few different websites that you can go to daily. Don’t pay for instruction over the In-ternet. Some websites require monthly or yearly fees but they aren’t worth it.Learn from a teacher. If it's possible,you should try to learn from as many different teachersas possible,but at the very least find one teacher and learn from that person.56.What is the passage mainly about?A. Why learning English is important.B. Some widely spoken languages.C. How to chat on line in English.D. How to learn English well.57. The author thinks reading English one houra day______-.A. is not easy but helpfulB. is interesting and helpfulC. can be stood by most peopleD. doesn’t help enlarge our vocabulary much58. What does the author suggest we do in the last but one paragraph?A. Pay for some instruction over the Internet.B. Find a good teacher to learn English.C. Find some free websites for Englishlearning.D. Learn English on three different websites each day.59. The passage is written for ______.A. housewivesB. language learnersC. businessmanD. travelersBWatson entered Mr. Smith's office. The boss was a hard man. He fired people who didn't do well without giving them a second chance. "Watson," said Mr. Smith, "this past year your department hasn't earned money. We're going to drop that department. It's finished. I’m sorry, —but you'll have to go." "But, sir —if I just had a little more time. For the moment I need the job to keep my son at Riverside School." "What's that!" said the boss. "Riverside! I didn't know you had a boy there. That's an expensive school for a man with your salary." "I know, sir. But he likes it there so much! He's a star athlete and the best boxer in the school. The boys call him Champ there."The boss sat perfectly still for a long time —a faraway (恍惚的) look in his eyes. Then, suddenly, he said, "We've got to close your department, Watson. But you'll take over a new job in another department. It means longer hours—maybe more pay. Now get out. You're here for life."Watson got out, with surprise on his face. Then the boss took a letter from the top drawer of his desk. It was Herbie's last letter from Riverside School —written a few days before he died. He had read it over and over again with sick pain. The letter read:I can’t say the boys here are any nicer to me than the others were. I guess it's the same everywhere when you're a cripple (跛脚的人). But don't worry about me, Dad. They've got a good chemistry department here. And there’s one boy here who is really great. He's a track star and boxing champ (冠军) and just tops in chemistry. The boys call him Champ. He made them stop throwing my books around. And he knocked a boydown who hit me. He is the best friend I ever had. Dad, when I grow up, 1 want to do something for Champ. Something big —that he won’ t even know about.Your son,Herbie60. Mr. Smith wanted to fire Watson because _____.A. Watson would take over a new job in another departmentB. Watson had his son study in Riverside SchoolC. Watson's son knocked a boy down who hit HerbieD. Watson's department didn't earn money that year61. We can learn from the text that _____.A. Mr. Smith didn't know that Champ was Watson's son at firstB. Mr. Smith was told not to fire Watson by HerbieC. Mr. Smith decided to give Watson another chance in no timeD. Mr. Smith wanted to realize Champ's dream62. According to the text, which words can best describe Mr. Smith?A. Selfish and greedy.B. Sympathetic and grateful.C. Hardworking and strict.D. Easy-going and optimistic.63. The author wrote the text in order to _____.A. share a moving story with usB. make our children enjoy lifeC. provide us with tips on workD. help us to deal with our bossCWho is behind wiseGEEK? It’s one of the most common questions we receive. We are a team of researchers, writers and editors providing short, clear and concise answers to common questions. Currently, there are over 200 active contributors; you can read some of their profiles (档案) below.64. What can we safely infer about wiseGEEK according to the profiles?A. It employed its staff members since the beginning of 2006.B. It is a website, offering brief and clear answers to common questions.C. It offers online news to help learn things one never knows existed.D. It aims at bringing up the curiosity of the children worldwide.65. _______ would be in charge if readers failed to visit wiseGEEK?A. CarolynB. BronwynC. JohnnyD. Catherine66. ______ of them do(es) extra work apart from working for wiseGEEK?A. OneB. TwoC. ThreeD.Four67. Which of the following is true according to the passage above?A. Carolyn really enjoys editing breaking news in early mornings.B. Bronwyn likes her tiger more than writing for wiseGEEK.C. Johnny holds the view that wiseGEEK enjoys great popularity.D. Catherine works for wiseGEEK mainly to travel in Europe.DThe word proactivity is fairly common in management literature, but you won't find it in the dictionary. It means that as a human being you take responsibility for your own life.Look at the word responsibility: ability to choose your response, response-ability. Effective people are proactive because they take responsibility. Their behavior is a product of their own decisions, based on values, rather than being a product of their own conditions, basedon feelings.For instance, you are planning a picnic with your family.You're excited.You have all the preparations.You've decided where to go, and then it becomes stormy, killing your plan.Proactive people carry weather within them. They realize what their purpose really was, and they creatively have a picnic elsewhere even if it's in their own basement with some special games, and make the best of that situation.The opposite of being proactive is to be reactive.Reactive people would say, "What's the use?" "We can't do anything." "Oh this is so upsetting after all of our preparations and arrangements." They try to persuade the people around them and usually the picnic will be cancelled.Being proactive is really just being true to your human nature. Your basic nature is to act, and not to be acted upon.That's true, despite widely accepted theories of determinism used to explain human nature. Determinism says that you don't really choose anything and that whatyou call choices are nothing more than automatic responses to outside conditions.The language of reactive people is like: " I can't." " Don't have time." " I have to." " I must." The whole spirit of that language is the transfer of responsibility.They think things are determined by their environment, or by their conditions, or by their conditioning or their genetic makeup.Psychologically, people who believe they are determined will produce the evidence to support the belief, and they increasingly feel victimized and out of control.They're not in charge of their life at all.On the contrary, a proactive person exercises free will, the freedom to choose the response that best applies to his values.In that way, he gains control over the circumstances, rather than being controlled by them.68.According to the passage, a proactive person's behavior can result from ______.A.the environmentB.an inner beliefC.the genetic makeupD.a temporary feeling 69.When a picnic plan is threatened by a sudden storm, reactive people will probably ____.A.have the picnic as plannedB.make the best of the picnic C.complain and give up the picnicD.find somewhere else for the picnic 70.What does "carry weather within them" in the second paragraph probably mean? A.Manage to improve the weather.B.Give in to the weather passively.C.Stress the influence of the weather.D. Find a solution to the weather problems.71.It can be concluded from the passage that determinists ______.A. accept things passivelyB. are in charge of themselvesC.are similar to proactive peopleD. respond to outside conditions activelyEShoppers who have cut back on purchases in the worrying U.S. economy are likely to continue their conservative shopping habits even if the economy improves, according to a new study.That signals continued bad news for retailers(零售商), who have been struggling to respond to a great cutback in spending as consumers have opened their wallets only for basic items, while cutting back on unimportant purchases.The report, which estimates retail sales growth to be flat this year, found that three-quarters of respondents to a monthly survey of 4,000 consumers said they had shifted their shopping behaviors because of theeconomy. Most said they were making do with less or going without some favorite items.“The habits learned during this economic crisis have the potential to permanently change the mind-set of consumers,” the report said. “The vast majority of shoppers who are changing their near-term shopping behavior say they plan to continue them as the economy improves.”To ease the consumer discomfort, smart retailers will focus on downsizing, the report said, as adding stores to an already crowded retail landscape is out of fashion. It came up with smaller initiatives to build shopping buzz, such as limited editions of products or having existing stores target local markets. The report also forecast that retailers will increasingly focus on private brands to attract cost-conscious shoppers, while adding complementary categories of goods to their stores to make one-stop shops for consumers.While warehouse clubs and super centers are expected to fare the best in the weak environment in 2009, discount department stores and supermarketswill be the weakest, the report warned. Sales of clothes, though expected to rebound in 2010, will be the hardest hit and slowest to recover.72 The current shopping behavior in the US is becoming________.A. arbitraryB. protectiveC. conservativeD. aggressive73. Which is the measure the retailers won’t take to attract shoppers?A. Adding more stores to the busy retail landscape.B. Keeping the editions of products limited.C. Making “one-stop” shops for customers.D. Concentrating on private brands and downsizing.74. What will the consumers do with the worrying US economy?A. Having existing stores target local markets.B. Going without some favorite items.C. Giving up their near-term shopping plan.D. Opening their wallets for daily buying.75. The underlined phrase “fare the best”most probably means “________”.A. setting the best exampleB. working out the bestC. making the greatest effortsD. having the most discount第Ⅱ卷(非选择题共25分)四.任务型读写(共10小题,每小题1分,满分10分)A conflict at work is common. If you can avoid conflict, it means you will win what you want regardless of what the other person wants. Since the potential issue has not been removed, it will simply reappear later. Let’s see what you can do.★ Be aware of the fact that some conflicts are unavoidable at work. On numerous occasions, conflict and disagreement are likely to happen. But when a conflict happens it’s notthe end of the world. On the contrary, it can be the beginning of an interesting learning process. Conflicts mean that people care enough to disagree strongly. The trick is not to allow the conflict to go on forever.★Deal with conflicts sooner rather than later. Solve a conflict when it starts, as it only gets worse with time going by. Conflicts at work arise not from something that was said, but from something that wasn’t said! Everyone’s waiting for the other to admit he’s wrong and gets more unpleasant after the conflict has lasted a while. It’s essential to interrupt the “waiting game” before i t gets to that point.★Ask nicely. If somebody has done something that made you angry, or if you don’t understand their viewpoint or actions, simply asking nicely about it can make a world of difference. Never assume that people do what they do to annoy o r hurt you. Sometimes there’s a good reason why that person does what he or she does, and a potential conflict disappear rightthere. Do remember to make an inquiry, not an accusation of any sort.★ Appreciate. Praise the other part in the conflict. Tell t hem why it’s worth it to you to solve the conflict. This can be difficult as few people find it easy to praise and appreciate a person they disagree strongly with, but it’s a great way to move forward.五.书面表达(满分15分)Rose是一位年轻的科学家,她每天工作繁忙且卓有成效。

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